A331425 -id:A331425 - cp's OEIS frontend

A331423 Divide each side of a triangle into n>=1 equal parts and trace the corresponding cevians, i.e., join every point, except for the first and last ones, with the opposite vertex. a(n) is the number of points at which three cevians meet.

0, 1, 0, 7, 0, 13, 0, 19, 0, 25, 0, 31, 0, 37, 6, 43, 0, 49, 0, 61, 0, 61, 0, 91, 0, 73, 0, 79, 0, 91, 0, 91, 0, 97, 12, 103, 0, 109, 0, 133, 0, 133, 0, 127, 42, 133, 0, 187, 0, 145, 0, 151, 0, 157, 12, 175, 0, 169, 0, 235, 0, 181, 48, 187, 6, 205, 0, 199, 0

Offset: 1

César Eliud Lozada, Jan 16 2020

Denote the cevians by a0, a1,...,an, b0, b1,...,bn, c0, c1,...,cn. For any given n, the indices (i,j,k) of (ai, bj, ck) meeting at a point are the integer solutions of:
n^3 - (i + j + k)*n^2 + (j*k + k*i + i*j)*n - 2*i*j*k = 0,  with 0 < i, j, k < n
or, equivalently and shorter,
(n-i)*(n-j)*(n-k) - i*j*k = 0, with 0 < i, j, k < n.
From N. J. A. Sloane, Feb 14 2020: (Start)
Stated another way, a(n) = number of triples (i,j,k) in [1,n-1] X [1,n-1] X [1,n-1] such that (i/(n-i))*(j/(n-j))*(k/(n-k)) = 1.
This is the quantity N3 mentioned in A091908.
Indices of zeros are precisely all odd numbers except those listed in A332378.
(End)

Antti Karttunen, Table of n, a(n) for n = 1..1045 (first 200 terms from Robert Israel)
Peter Kagey, An illustration of A331423(4) = 7.
Hugo Pfoertner, Visualization of diagonal intersections in an equilateral triangle.
Jim Propp and Adam Propp-Gubin, Counting Triangles in Triangles, arXiv:2409.17117 [math.CO], 2024. See p. 9.

Cf. A091908, A332378. Bisections are A331425, A331428.

Ceva:= proc(n) local a, i, j, k; a:=0;
for i from 1 to n-1 do
for j from 1 to n-1 do
for k from 1 to n-1 do
if i*j*k/((n-i)*(n-j)*(n-k)) = 1 then a:=a+1; fi;
od: od: od: a; end;
t1:=[seq(Ceva(n), n=1..80)];  # N. J. A. Sloane, Feb 14 2020

CevIntersections[n_] := Length[Solve[(n - i)*(n - j)*(n - k) - i*j*k == 0 && 0 < i < n &&  0 < j < n && 0 < k < n, {i, j, k}, Integers]];
Map[CevIntersections[#] &, Range[50]]

A331423(n) = sum(i=1, n-1, sum(j=1, n-1, sum(k=1, n-1, (1==(i*j*k)/((n-i)*(n-j)*(n-k)))))); \\ (After the Maple program) - Antti Karttunen, Dec 12 2021

A331428 Divide each side of a triangle into 2*n-1 (n>=1) equal parts and trace the corresponding cevians, i.e., join every point, except for the first and last ones, with the opposite vertex. a(n) is the number of points at which three cevians meet.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12, 0, 0, 0, 0, 42, 0, 0, 0, 0, 12, 0, 0, 0, 48, 6, 0, 0, 0, 0, 6, 12, 0, 0, 0, 6, 0, 0, 12, 0, 0, 0, 24, 0, 0, 90, 0, 0, 0, 0, 0, 12, 12, 0, 0, 0, 0, 0, 0, 0, 66, 0, 0, 0, 24, 0, 0, 0, 0, 12, 0, 0, 0, 12, 0

Offset: 1

César Eliud Lozada, Jan 16 2020

nonn

A bisection of A331423.

Cf. A331423, A331425.

CevIntersections[n_] := Length[Solve[(n - i)*(n - j)*(n - k) - i*j*k == 0 && 0 < i < n &&  0 < j < n && 0 < k < n, {i, j, k}, Integers]];
Map[CevIntersections[#] &, Range[1,51,2]]

a(n) = A331423(2*n-1).

cp's OEIS Frontend

A331423 Divide each side of a triangle into n>=1 equal parts and trace the corresponding cevians, i.e., join every point, except for the first and last ones, with the opposite vertex. a(n) is the number of points at which three cevians meet.

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